# I When to use which dimensionless number

1. Dec 15, 2017

### Staff: Mentor

I get $$\frac{\sqrt{g_{zz}}}{\sqrt{g_{zz}g_{\theta \theta}-g_{\theta z}^2}}$$

2. Dec 16, 2017

### joshmccraney

Ok, so we have:
$$\hat{a}^\theta = \frac{g_{zz}}{-g_{\theta z}^2+g_{zz}g_{\theta\theta}}\hat{a}_\theta+\frac{g_{\theta z}}{g_{\theta z}^2-g_{zz}g_{\theta\theta}}\hat{a}_z \implies\\ |\hat{a}^\theta| = \sqrt{\left(\frac{g_{zz}}{-g_{\theta z}^2+g_{zz}g_{\theta\theta}}\right)^2+\left(\frac{g_{\theta z}}{g_{\theta z}^2-g_{zz}g_{\theta\theta}}\right)^2}\\ =\sqrt{\frac{g_{zz}^2}{(-g_{\theta z}^2+g_{zz}g_{\theta\theta})^2}+\frac{g_{\theta z}^2}{(g_{\theta z}^2-g_{zz}g_{\theta\theta})^2}}\\ =\sqrt{\frac{g_{zz}^2+g_{\theta z}^2} {(-g_{\theta z}^2+g_{zz}g_{\theta\theta})^2}}$$
What do you think?

3. Dec 16, 2017

### Staff: Mentor

I think that $\hat{a}_{\theta}$ and $\hat{a}_{z}$ are not orthogonal, nor are they unit vectors.

4. Dec 16, 2017

### joshmccraney

Riiiiiiight, as you said bevore $$\hat{a}_\theta=\frac{\partial R}{\partial \theta}\hat{r}+R\hat{\theta}\tag{3a}$$and$$\hat{a}_z=\frac{\partial R}{\partial z}\hat{r}+\hat{z}\tag{3b}$$
My bad. So I would take the norm by transforming back to these definitions. Somehow I have a feeling you have a more general approach? Could you share it and I'll do the footwork? I'd like to see how to take a vector norm with non-orthonormal basis vectors.

5. Dec 16, 2017

### Staff: Mentor

Just use Eqns. 5 in post #35.

6. Jan 3, 2018

### joshmccraney

I tried thinking about how equations 5 help, but I can't see how. At any rate, we have
$$\hat{a}_\theta=\frac{\partial R}{\partial \theta}\hat{r}+R\hat{\theta}\tag{3a}$$
$$\hat{a}_z=\frac{\partial R}{\partial z}\hat{r}+\hat{z}\tag{3b}$$
$$\hat{a}^\theta = \frac{g_{zz}}{-g_{\theta z}^2+g_{zz}g_{\theta\theta}}\hat{a}_\theta+\frac{g_{\theta z}}{g_{\theta z}^2-g_{zz}g_{\theta\theta}}\hat{a}_z \\ =\frac{-g_{zz}}{g_{\theta z}^2-g_{zz}g_{\theta\theta}} \left( \frac{\partial R}{\partial \theta}\hat{r}+R\hat{\theta} \right)+\frac{g_{\theta z}}{g_{\theta z}^2-g_{zz}g_{\theta\theta}}\left( \frac{\partial R}{\partial z}\hat{r}+\hat{z} \right)\\ = \frac{g_{\theta z} \partial_zR -g_{zz} \partial_\theta R}{g_{\theta z}^2-g_{zz}g_{\theta\theta}} \hat{r} + \frac{-g_{zz} R}{g_{\theta z}^2-g_{zz}g_{\theta\theta}} \hat{\theta} + \frac{g_{\theta z} }{g_{\theta z}^2-g_{zz}g_{\theta\theta}} \hat{z}$$
where from here I could compute the norm as a regular problem. Would you mind highlighting the procedure you referenced, involving equations 5? Again, I'm sorry for the amount of time it's taken for me to reply!

7. Jan 3, 2018

### Staff: Mentor

$$\hat{a}^{\theta}=\frac{g_{zz}\hat{a}_{\theta}-g_{\theta z}\hat{a}_z}{g_{zz}g_{\theta \theta}-g_{\theta z}^2}$$
$$\hat{a}^{\theta}\centerdot \hat{a}^{\theta}=\frac{(g_{zz}\hat{a}_{\theta}-g_{\theta z}\hat{a}_z)\centerdot (g_{zz}\hat{a}_{\theta}-g_{\theta z}\hat{a}_z)}{(g_{zz}g_{\theta \theta}-g_{\theta z}^2)^2}$$
$$\hat{a}^{\theta}\centerdot \hat{a}^{\theta}=\frac{g_{zz}^2(\hat{a}_{\theta}\centerdot \hat{a}_{\theta})-2g_{\theta z}g_{zz}(\hat{a}_z\centerdot \hat{a}_{\theta})+g_{\theta z}^2(\hat{a}_z\centerdot \hat{a}_z)}{(g_{zz}g_{\theta \theta}-g_{\theta z}^2)^2}$$
$$\hat{a}^{\theta}\centerdot \hat{a}^{\theta}=\frac{g_{zz}^2g_{\theta \theta}-2g_{\theta z}^2g_{zz}+g_{\theta z}^2g_{zz}}{(g_{zz}g_{\theta \theta}-g_{\theta z}^2)^2}$$
$$\hat{a}^{\theta}\centerdot \hat{a}^{\theta}=\frac{g_{zz}(g_{zz}g_{\theta \theta}-g_{\theta z}^2)}{(g_{zz}g_{\theta \theta}-g_{\theta z}^2)^2}$$
$$\hat{a}^{\theta}\centerdot \hat{a}^{\theta}=\frac{g_{zz}}{g_{zz}g_{\theta \theta}-g_{\theta z}^2}$$

8. Jan 3, 2018

### joshmccraney

Riiiiiight, the magnitude of any vector $\vec v =\sqrt{ \vec v \cdot \vec v}$ since $\vec v \cdot \vec v = |v| |v| \cos \theta = |v|^2$. Shoot, this makes perfect sense now! Okay, so I agree the magnitude is $$\frac{\sqrt{g_{zz}}}{\sqrt{g_{zz}g_{\theta \theta}-g_{\theta z}^2}}.$$ From the above we know
$$\hat{a}^\theta = \frac{g_{zz}}{-g_{\theta z}^2+g_{zz}g_{\theta\theta}}\hat{a}_\theta+\frac{g_{\theta z}}{g_{\theta z}^2-g_{zz}g_{\theta\theta}}\hat{a}_z\implies\\ \hat u^\theta = \sqrt{g_{zz}}\hat{a}_\theta - \frac{g_{\theta z}}{\sqrt{g_{zz}}}\hat{a}_z\\ =\frac{g_{zz}\hat{a}_\theta - g_{\theta z}\hat{a}_z} {\sqrt{g_{zz}}}.$$
How does that look?

9. Jan 3, 2018

### Staff: Mentor

I think you already had the equation for $\hat{a}^{\theta}$. If you're trying to represent the unit vector in the same direction as $\hat{a}^{\theta}$ (which this is not), then at you're missing a factor of $\sqrt{g_{zz}g_{\theta \theta}-g_{\theta z}^2}$ in the denominator.

10. Jan 3, 2018

### joshmccraney

You previously posted